Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 26

$\frac{1}{4}tan^8(\frac{x}{2})+C$

Let u=$tan(\frac{x}{2})=>du=\frac{1}{2}sec^2(\frac{x}{2})dx=>2du=sec^2(\frac{x}{2})dx$ =$\int tan^7(\frac{x}{2})sec^(\frac{x}{2})dx$ =$\int u^7(2du)=2(\frac{1}{8}u^8)+C$ =$\frac{1}{4}tan^8(\frac{x}{2})+C$